Revisal Problems (Past 13 Years) JEE Main (Chemical Bonding) - JEE MCQ

has three covalent and one coordinate bond.

Species with odd number of electrons will certainly have unpaired electron in bonding, thus paramagnetic.

Thus, OH, NO₃ and HO₂ are paramagnetic.

Bonding in metals is called metallic bonding. It results from the electrical attractions among positively charged metal ions and mobile, delocalised electrons belonging to the crystal as a whole. Two theories:
(a) Band metal
(b) Electron-sea model

 l-atom is sp³d³ -hybridised

Bond angle = 72° {for five) and = 90° (for two)
No lone pair

This is the structure for AsF5 which is Trigonal Bipyramidal, it has sp3d hybridization.

So, s and all three orbitals of p will be used. Since the bonds with the remaining two Fs are in the perpendicular plane, when we compare it to the shape of different d- orbitals, we get dz2 as the answer.

Hence C is the correct answer.

Valence electrons = 5

Each P-atom has three (P— P) bonds and one lone pair and is sp3-hybridised.

Thus, two σ and two π-bonds.

There are five bonding pairs or electron pairs. Thus, sp³d-hybridisation. Trigonal bipyramidal structure.
Three bonding pairs are at 120° with one each other.
Two bonding pairs are at 90° with the plane in which first three bonding pairs exist.

Intermolecular H-bonding (dimer is formed).


Intramolecular H-bonding (makes it more volatile than m-and p-isomers)

Intermolecular H-bonding.
IV. H₃BO₃ : Hexagonal structure is formed due to intermolecular H-bonding.

(a) Chelation is ring formation.



Intramolecular H-bonding.

Thus, HF > H₂S > H₂O

Going down the group, size of the cation increases. Since, the lattice energy decreases much more than the hydration energy with increasing size solubility of hydroxide increases (when size of cation and anion are of comparable size). (b) and (c) If the anion is large com p are d to cation, the lattice energy will remain almost constant. Since, hydration energy decreases down a group, solubility will decrease hence (b) and (c) are incorrect.

KO₂ has superoxide  Thus, one electron is upaired.

Compound is said to be ionic if (EN) difference between bonding atoms exceeds 1.7.

(Note CsF is most ionic.}

H₂O and F^- are weak ligands. Hence, unpaired electrons in Fe³⁺ is (I) and (IV) are not affected. Thus outer d-orbitals are involved in hybridisation.

Six sp³d²-orbitals accept lone pair from weak ligands - octahedral geometry.

There are (Cl → O) coordinate bonds. Thus, bond length is shorter than the normal (Cl—O) bond length (1.72A°). Thus, there is considerable double bond character. Thus, both statements I and II are correct and statement II is the correct explanation statement I.

N₂O₄ is formed by dimerisation of NO₂.

Unpaired electron is shared and (N—N) bond is formed. This results is formation of on each N-atom. Thus, dipole-dipole repulsion increases (N—N) bond length (larger than its normal value . Thus, both statements I and II are correct but statement II is not the correct explanation of statement I.

There are two bond pairs and three lone pairs in the outer shell of central atom. To minimise the repulsive forces, the three lone pairs occupy the equatorial position. The ion is therefore linear in shape with a bond angle of exactly 180°.

Thus, both statements I and II are correct but statement II is not the correct explanation of statement I,

and are of same group. I can expand its octet but F cannot expand its octet, thus in octet rule is not violated but in , octet rule is violated.
F^- (electron rich) is a Lewis base.
Thus, statement I is correct and statement II is incorrect.